8: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

a) 3x-6=0

3x=6 => x=2

b) (3x+2)(4x-5)=0

=> 3x+2=0 => x=-2/3

hoặc 4x-5=0 => x=5/4

câu c ,d thiếu dấu '=" để thành 1 pt rồi bạn

c) \(\dfrac{2x-5}{3}+\dfrac{x-3}{5}=\dfrac{4x+3}{15}\)

=> 10x -25 +3X-9=4X+3

=>9x=37

=>x=37/9

d) \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\) ĐK (x khác 3,-3)

=>5x+15+4x-12=x-5

=>8x=-8

=>x=-1

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a) Đặt x^2+2x+2=t

\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)

\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)

Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

bạn viết rõ đề câu a;b nhé 

c, \(2x\left(x-5\right)-\left(x-5\right)=0\Leftrightarrow\left(2x-1\right)\left(x-5\right)=0\Leftrightarrow x=\dfrac{1}{2};x=5\)

d, \(\left(x+3\right)\left(x+3-5+x\right)=0\Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\Leftrightarrow x=-3;x=1\)

e, \(\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{5}\)

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4c^3\right)\)

\(=-\dfrac{1}{2}\left(2x+6-4c^3\right)+3\left(2x+6-4c^3\right)\)

\(=-x^2-3x+2c^3x+6x+18-12c^3\)

\(=-x^2+3x+2c^3x+18-12c^3\)

f) \(\left(2x-5\right)\left(x^2-x+3\right)\)

\(=2x\left(x^2-x+3\right)-5\left(x^2-x+3\right)\)

\(=2x^3-2x^2+6x-5x^2+5x-15\)

\(=2x^3-7x^2+11x-15\)

w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)

\(=3x\left(x^2-2x-5\right)+\left(x^2-2x-5\right)\)

\(=3x^3-6x^2-15x+x^2-2x-5\)

\(=3x^3-5x^2-17x-5\)

x) \(\left(6x-3\right)\left(x^2+x-1\right)\)

\(=6x\left(x^2+x-1\right)-3\left(x^2+x-1\right)\)

\(=6x^3+6x^2-6x-3x^2-3x+3\)

\(=6x^3+3x^2-9x+3\)

y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)

\(=5x\left(3x+1-x^2\right)-2\left(3x+1-x^2\right)\)

\(=15x^2+5x-5x^3-6x-2+2x^2\)

\(=-5x^3+17x^2-x-2\)

z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)

\(=\dfrac{3}{4}x\left(4x^2+4x+4\right)+\left(4x^2+4x+4\right)\)

\(=3x^3+3x^2+3x+4x^2+4x+4\)

\(=3x^3+7x^2+7x+4\)

f: =2x^3-2x^2+6x-5x^2+5x-15

=2x^3-7x^2+11x-15

w: =3x^3-6x^2-15x+x^2-2x-5

=3x^3-5x^2-17x-5

x: =6x^3+6x^2-6x-3x^2-3x+3

=6x^3+3x^2-9x+3

y: =(5x-2)(-x^2+3x+1)

=-5x^3+15x^2+5x+2x^2-6x-2

=-5x^3+17x^2-x-2

z: =3x^3+3x^2+3x+4x^2+4x+4

=3x^3+7x^2+7x+4

ai giúp mình với

plz

tham khảo

f(x) = x5 + 3x2 − 5x3 − x7 + x3 + 2x2 + x5 − 4x2 + x7

= (x5 + x5) + (3x2 + 2x2 – 4x2) + (-5x3 + x3) + (-x7 + x7)

= 2x5 + x2 – 4x3.

= 2x5 - 4x3 + x2

Đa thức có bậc là 5

g(x) = x4 + 4x3 – 5x8 – x7 + x3 + x2 – 2x7 + x4 – 4x2 – x8

= (x4 + x4) + (4x3 + x3) – (5x8 + x8) – (x7 + 2x7) + (x2 – 4x2)

= 2x4 + 5x3 – 6x8 – 3x7 – 3x2

= -6x8 - 3x7 + 2x4 + 5x3 - 3x2.

Đa thức có bậc là 8.

1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)

\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)

\(=\dfrac{1}{2}x^3+x^2-15x-18\)

2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)

\(=4x^3+6x^2-6x^2-9x+10x+15\)

\(=4x^3+x+15\)

3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)

\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)

\(=3x^5-x^4+5x^3+10x^2+26x-5\)

4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)

\(=\left(x^2-1\right)\left(x-2\right)\)

\(=x^3-2x^2-x+2\)